Term Rewriting System R:
[x, y]
g(c(x, s(y))) -> g(c(s(x), y))
f(c(s(x), y)) -> f(c(x, s(y)))
f(f(x)) -> f(d(f(x)))
f(x) -> x

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

G(c(x, s(y))) -> G(c(s(x), y))
F(c(s(x), y)) -> F(c(x, s(y)))
F(f(x)) -> F(d(f(x)))

Furthermore, R contains two SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Instantiation Transformation
       →DP Problem 2
         ↳Inst


Dependency Pair:

G(c(x, s(y))) -> G(c(s(x), y))


Rules:


g(c(x, s(y))) -> g(c(s(x), y))
f(c(s(x), y)) -> f(c(x, s(y)))
f(f(x)) -> f(d(f(x)))
f(x) -> x


Strategy:

innermost




On this DP problem, an Instantiation SCC transformation can be performed.
As a result of transforming the rule

G(c(x, s(y))) -> G(c(s(x), y))
one new Dependency Pair is created:

G(c(s(x''), s(y''))) -> G(c(s(s(x'')), y''))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Inst
           →DP Problem 3
             ↳Instantiation Transformation
       →DP Problem 2
         ↳Inst


Dependency Pair:

G(c(s(x''), s(y''))) -> G(c(s(s(x'')), y''))


Rules:


g(c(x, s(y))) -> g(c(s(x), y))
f(c(s(x), y)) -> f(c(x, s(y)))
f(f(x)) -> f(d(f(x)))
f(x) -> x


Strategy:

innermost




On this DP problem, an Instantiation SCC transformation can be performed.
As a result of transforming the rule

G(c(s(x''), s(y''))) -> G(c(s(s(x'')), y''))
one new Dependency Pair is created:

G(c(s(s(x'''')), s(y''''))) -> G(c(s(s(s(x''''))), y''''))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Inst
           →DP Problem 3
             ↳Inst
             ...
               →DP Problem 4
                 ↳Argument Filtering and Ordering
       →DP Problem 2
         ↳Inst


Dependency Pair:

G(c(s(s(x'''')), s(y''''))) -> G(c(s(s(s(x''''))), y''''))


Rules:


g(c(x, s(y))) -> g(c(s(x), y))
f(c(s(x), y)) -> f(c(x, s(y)))
f(f(x)) -> f(d(f(x)))
f(x) -> x


Strategy:

innermost




The following dependency pair can be strictly oriented:

G(c(s(s(x'''')), s(y''''))) -> G(c(s(s(s(x''''))), y''''))


There are no usable rules for innermost that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

resulting in one new DP problem.
Used Argument Filtering System:
G(x1) -> G(x1)
c(x1, x2) -> x2
s(x1) -> s(x1)


   R
     ↳DPs
       →DP Problem 1
         ↳Inst
           →DP Problem 3
             ↳Inst
             ...
               →DP Problem 5
                 ↳Dependency Graph
       →DP Problem 2
         ↳Inst


Dependency Pair:


Rules:


g(c(x, s(y))) -> g(c(s(x), y))
f(c(s(x), y)) -> f(c(x, s(y)))
f(f(x)) -> f(d(f(x)))
f(x) -> x


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Inst
       →DP Problem 2
         ↳Instantiation Transformation


Dependency Pair:

F(c(s(x), y)) -> F(c(x, s(y)))


Rules:


g(c(x, s(y))) -> g(c(s(x), y))
f(c(s(x), y)) -> f(c(x, s(y)))
f(f(x)) -> f(d(f(x)))
f(x) -> x


Strategy:

innermost




On this DP problem, an Instantiation SCC transformation can be performed.
As a result of transforming the rule

F(c(s(x), y)) -> F(c(x, s(y)))
one new Dependency Pair is created:

F(c(s(x''), s(y''))) -> F(c(x'', s(s(y''))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Inst
       →DP Problem 2
         ↳Inst
           →DP Problem 6
             ↳Instantiation Transformation


Dependency Pair:

F(c(s(x''), s(y''))) -> F(c(x'', s(s(y''))))


Rules:


g(c(x, s(y))) -> g(c(s(x), y))
f(c(s(x), y)) -> f(c(x, s(y)))
f(f(x)) -> f(d(f(x)))
f(x) -> x


Strategy:

innermost




On this DP problem, an Instantiation SCC transformation can be performed.
As a result of transforming the rule

F(c(s(x''), s(y''))) -> F(c(x'', s(s(y''))))
one new Dependency Pair is created:

F(c(s(x''''), s(s(y'''')))) -> F(c(x'''', s(s(s(y'''')))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Inst
       →DP Problem 2
         ↳Inst
           →DP Problem 6
             ↳Inst
             ...
               →DP Problem 7
                 ↳Argument Filtering and Ordering


Dependency Pair:

F(c(s(x''''), s(s(y'''')))) -> F(c(x'''', s(s(s(y'''')))))


Rules:


g(c(x, s(y))) -> g(c(s(x), y))
f(c(s(x), y)) -> f(c(x, s(y)))
f(f(x)) -> f(d(f(x)))
f(x) -> x


Strategy:

innermost




The following dependency pair can be strictly oriented:

F(c(s(x''''), s(s(y'''')))) -> F(c(x'''', s(s(s(y'''')))))


There are no usable rules for innermost that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
c > s

resulting in one new DP problem.
Used Argument Filtering System:
F(x1) -> F(x1)
c(x1, x2) -> c(x1, x2)
s(x1) -> s(x1)


   R
     ↳DPs
       →DP Problem 1
         ↳Inst
       →DP Problem 2
         ↳Inst
           →DP Problem 6
             ↳Inst
             ...
               →DP Problem 8
                 ↳Dependency Graph


Dependency Pair:


Rules:


g(c(x, s(y))) -> g(c(s(x), y))
f(c(s(x), y)) -> f(c(x, s(y)))
f(f(x)) -> f(d(f(x)))
f(x) -> x


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:00 minutes